What is the integral of (x^2+1)/(x^3+x^2-2x) ?

The integral of (x^2+1)/(x^3+x^2-2x) is -1/2 ln|x|+2/3 ln|x-1|+5/6 ln|x+2|+C

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integral of (x^2+1)/(x^3+x^2-2x)

\int \frac{x ^{2} + 1}{x ^{3} + x ^{2} - 2 x} d x

- \frac{1}{2} ln ∣ x ∣ + \frac{2}{3} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{x ^{2} + 1}{x ^{3} + x ^{2} - 2 x} d x

Take the partial fraction of

\frac{x ^{2} + 1}{x ^{3} + x ^{2} - 2 x} : - \frac{1}{2 x} + \frac{2}{3 ( x - 1 )} + \frac{5}{6 ( x + 2 )}

= \int - \frac{1}{2 x} + \frac{2}{3 ( x - 1 )} + \frac{5}{6 ( x + 2 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{2 x} d x + \int \frac{2}{3 ( x - 1 )} d x + \int \frac{5}{6 ( x + 2 )} d x

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

\int \frac{2}{3 ( x - 1 )} d x = \frac{2}{3} ln ∣ x - 1 ∣

\int \frac{5}{6 ( x + 2 )} d x = \frac{5}{6} ln ∣ x + 2 ∣

= - \frac{1}{2} ln ∣ x ∣ + \frac{2}{3} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 2 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x ∣ + \frac{2}{3} ln ∣ x - 1 ∣ + \frac{5}{6} ln ∣ x + 2 ∣ + C

\int \frac{e ^{2 x}}{e ^{2 x}} d x

f (x) = \frac{( 4 x ^{5} + 8 )}{x ^{3}}

f (x) = 9 x^{\frac{1}{3}}

y^{^{'}} 1 - 4 x^{2} = x

y^{4} y^{^{'}} = - 3 x^{2} y^{5} + x^{2}

What is the integral of (x^2+1)/(x^3+x^2-2x) ?
The integral of (x^2+1)/(x^3+x^2-2x) is -1/2 ln|x|+2/3 ln|x-1|+5/6 ln|x+2|+C